【HDOJ】3505 Writing Robot

挺好的一道题目，我的做法是kmp+Dinic网络流。
kmp求子串在P中出现的次数，从而计算love值。
网络流主要用来处理最优解。
case2中p1的love值是8，p2的love值是7，最终T包含p1和p2，hate值也仅仅算一次。
这个题目难点在于思考为什么网络流的解法是合理，可以反证。从而导出最优解等于love的总和-最大流。
建图方法：
source连接p，初始容量为love值；
p连接s，初始容量为INF;
s连接destination，容量为hate值。
在最优解中，如果s到des的流量小于容量，则证明包含s的p都不被选择。即减去p的容量和。如果流量大于等于容量，则证明该直接去掉hate值。

 /* 3505 */
 #include <iostream>
 #include <string>
 #include <map>
 #include <queue>
 #include <set>
 #include <stack>
 #include <vector>
 #include <deque>
 #include <algorithm>
 #include <cstdio>
 #include <cmath>
 #include <ctime>
 #include <cstring>
 #include <climits>
 #include <cctype>
 #include <cassert>
 #include <functional>
 #include <iterator>
 #include <iomanip>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,1024000")

 #define sti                set<int>
 #define stpii            set<pair<int, int> >
 #define mpii            map<int,int>
 #define vi                vector<int>
 #define pii                pair<int,int>
 #define vpii            vector<pair<int,int> >
 #define rep(i, a, n)     for (int i=a;i<n;++i)
 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 #define clr                clear
 #define pb                 push_back
 #define mp                 make_pair
 #define fir                first
 #define sec                second
 #define all(x)             (x).begin(),(x).end()
 #define SZ(x)             ((int)(x).size())
 #define lson            l, mid, rt<<1
 #define rson            mid+1, r, rt<<1|1

 typedef struct {
     int l, h, len;
     char s[];
     int nxt[];

     void getnext() {
         int i = , j = -;

         nxt[] = -;
         while (i < len) {
             if (j==- || s[i]==s[j]) {
                 ++i;
                 ++j;
                 nxt[i] = j;
             } else {
                 j = nxt[j];
             }
         }
     }

     int kmp(char *d) {
         int dlen = strlen(d);
         int i = , j =;
         int ret = ;

         while (i < dlen) {
             if (d[i] == s[j]) {
                 ++i;
                 ++j;
             } else {
                 j = nxt[j];
                 if (j == -) {
                     j = ;
                     ++i;
                 }
             }
             if (j == len) {
                 ++ret;
                 j = nxt[j];
             }
         }

         return ret;
     }

 } node_t;

 typedef struct {
     int v, f, nxt;
 } edge_t;

 const int INF = 0x3f3f3f3f;
 const int maxn = ;
 const int maxv = maxn * ;
 const int maxl = ;
 const int maxe = 1e5+;
 node_t nd[maxn];
 bool M[maxn][maxn];
 int score[maxn];
 int head[maxv], head_[maxv];
 edge_t E[maxe];
 int dis[maxv];
 int Q[maxv];
 char s[maxl];
 int nxt[maxl];
 int sn, pn, m;

 int calc(int k) {
     int ret = , cnt;

     rep(i, , sn+) {
         cnt = nd[i].kmp(s);
         if (cnt) {
             ret += nd[i].l * cnt;
             M[k][i] = true;
         }
     }

     return ret;
 }

 void init() {
     m = ;
     memset(head, -, sizeof(head));
     memset(M, false, sizeof(M));
 }

 void addEdge(int u, int v, int f) {
     E[m].v = v;
     E[m].f = f;
     E[m].nxt = head[u];
     head[u] = m++;

     E[m].v = u;
     E[m].f = ;
     E[m].nxt = head[v];
     head[v] = m++;
 }

 bool bfs(int s, int t) {
     int l = , r = ;
     int u, v, k;

     Q[r++] = s;
     memset(dis, , sizeof(dis));
     dis[s] = ;

     while (l < r) {
         u = Q[l++];
         for (k=head[u]; k!=-; k=E[k].nxt) {
             v = E[k].v;
             if (!dis[v] && E[k].f) {
                 dis[v] = dis[u] + ;
                 if (v == t)
                     return false;
                 Q[r++] = v;
             }
         }
     }

     return true;
 }

 int dfs(int u, int t, int val) {
     if (u==t || val==)
         return val;

     int ret = ;
     int v, tmp;

     for (int& k=head_[u]; k!=-; k=E[k].nxt) {
         v = E[k].v;
         if (E[k].f && dis[v]==dis[u]+ && (tmp=dfs(v, t, min(val, E[k].f)))>) {
             E[k].f -= tmp;
             E[k^].f += tmp;
             ret += tmp;
             val -= tmp;
             if (val == )
                 return ret;
         }
     }

     return ret;
 }

 int Dinic(int s, int t) {
     int ret = , tmp;

     while () {
         if (bfs(s, t))
             break;

         memcpy(head_, head, sizeof(head));
         while () {
             tmp = dfs(s, t, INF);
             if (tmp == )
                 break;
             ret += tmp;
         }
     }

     return ret;
 }

 void Build() {
     int s = ;
     int t = maxv - ;

     rep(i, , sn+) {
         addEdge(i, t, nd[i].h);
     }

     rep(i, , pn+) {
         addEdge(s, sn+i, score[i]);
         rep(j, , sn+) {
             if (M[i][j])
                 addEdge(sn+i, j, INF);
         }
     }
 }

 int solve() {
     int ret = ;

     rep(i, , pn+) {
         scanf("%s", s);
         score[i] = calc(i);
         ret += score[i];
     }

     Build();
     int tmp = Dinic(, maxv-);
     ret -= tmp;
     #ifndef ONLINE_JUDGE
         printf("maxflow = %d\n", tmp);
     #endif

     return ret;
 }

 int main() {
     ios::sync_with_stdio(false);
     #ifndef ONLINE_JUDGE
         freopen("data.in", "r", stdin);
         freopen("data.out", "w", stdout);
     #endif

     int t;
     int ans;

     scanf("%d", &t);
     rep(tt, , t+) {
         init();
         scanf("%d %d", &sn, &pn);
         rep(i, , sn+) {
             scanf("%d %d %s", &nd[i].l, &nd[i].h, nd[i].s);
             nd[i].len = strlen(nd[i].s);
             nd[i].getnext();
         }
         ans = solve();
         printf("Case %d: %d\n", tt, ans);
     }

     #ifndef ONLINE_JUDGE
         printf("time = %d.\n", (int)clock());
     #endif

     return ;
 }